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Hi Greg M.,
You are right that %path% is
interpreted differently in MSDOS 6.22.
That is why I said, "Don't ask me for
particulars", in a previous post. I knew
in the back of my mind that there was
something tricky involved. %path% is
only interpreted as the path statement
when it is used in a batch file in 6.22 .
From the command line it is interpreted
literally as "%path%".
Eric
On Mon, 27 Nov 2006 10:11:34 +1030 "Greg Mayman" <gmone@xxxxxxxxxx>
writes:
> Arachne at FreeLists---The Arachne Fan Club!
>
> On Sat, 25 Nov 2006 05:56:05 -0500, Eric S. Emerson wrote:
>
> > By using %path1%;%path2%;...... you can
> > make it longer.
>
> That's what I thought, but it doesn't work that way in MS-DOS 6.22.
>
> When I tried it in a batch file, I found that the values of path1
> and path2 were substituted and it wass executed as if it was
> written in full.
>
> Try it yourself with this batch file that doubles the length of
> the path statement:
>
> set old=%path%
> path %path%;%path%
> path %old%
>
> Note that echo is left on so you can see how the commands are
> executed.
>
> If you want to see the truncation, try this batch file:
>
> set old=%path%
> path %path%;%path%
> path
> pause
>
> then repeat the last three lines at least four times, and finish
> with
> path %old%
> to restore the original path.
>
> . ,-./\
> . / \ from Greg Mayman, in Adelaide, South Australia
> . \_,-*_/ "Queen City of The South" 34:55 S 138:36 E
> . v
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